Solve by Substitution y+16x-22=4x^2 4x^2-24x+26+y=0

$y + 16 x - 22 = 4 x^{2}$ $4 x^{2} - 24 x + 26 + y = 0$

Move all terms not containing $y$ to the right side of the equation.

Subtract $16 x$ from both sides of the equation.

$y - 22 = 4 x^{2} - 16 x$

$4 x^{2} - 24 x + 26 + y = 0$

Add $22$ to both sides of the equation.

$y = 4 x^{2} - 16 x + 22$

$4 x^{2} - 24 x + 26 + y = 0$

$y = 4 x^{2} - 16 x + 22$

$4 x^{2} - 24 x + 26 + y = 0$

Replace all occurrences of $y$ with $4 x^{2} - 16 x + 22$ in each equation.

Replace all occurrences of $y$ in $4 x^{2} - 24 x + 26 + y = 0$ with $4 x^{2} - 16 x + 22$ .

$4 x^{2} - 24 x + 26 + 4 x^{2} - 16 x + 22 = 0$

$y = 4 x^{2} - 16 x + 22$

Simplify the left side.

Simplify $4 x^{2} - 24 x + 26 + 4 x^{2} - 16 x + 22$ .

Remove parentheses.

$4 x^{2} - 24 x + 26 + 4 x^{2} - 16 x + 22 = 0$

$y = 4 x^{2} - 16 x + 22$

Add $4 x^{2}$ and $4 x^{2}$ .

$8 x^{2} - 24 x + 26 - 16 x + 22 = 0$

$y = 4 x^{2} - 16 x + 22$

Subtract $16 x$ from $- 24 x$ .

$8 x^{2} - 40 x + 26 + 22 = 0$

$y = 4 x^{2} - 16 x + 22$

Add $26$ and $22$ .

$8 x^{2} - 40 x + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

$8 x^{2} - 40 x + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

$8 x^{2} - 40 x + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

$8 x^{2} - 40 x + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

Solve for $x$ in $8 x^{2} - 40 x + 48 = 0$ .

Factor the left side of the equation.

Factor $8$ out of $8 x^{2} - 40 x + 48$ .

Factor $8$ out of $8 x^{2}$ .

$8 (x^{2}) - 40 x + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

Factor $8$ out of $- 40 x$ .

$8 (x^{2}) + 8 (- 5 x) + 48 = 0$

$y = 4 x^{2} - 16 x + 22$

Factor $8$ out of $48$ .

$8 x^{2} + 8 (- 5 x) + 8 \cdot 6 = 0$

$y = 4 x^{2} - 16 x + 22$

Factor $8$ out of $8 x^{2} + 8 (- 5 x)$ .

$8 (x^{2} - 5 x) + 8 \cdot 6 = 0$

$y = 4 x^{2} - 16 x + 22$

Factor $8$ out of $8 (x^{2} - 5 x) + 8 \cdot 6$ .

$8 (x^{2} - 5 x + 6) = 0$

$y = 4 x^{2} - 16 x + 22$

$8 (x^{2} - 5 x + 6) = 0$

$y = 4 x^{2} - 16 x + 22$

Factor.

Factor $x^{2} - 5 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

$y = 4 x^{2} - 16 x + 22$

Write the factored form using these integers.

$8 ((x - 3) (x - 2)) = 0$

$y = 4 x^{2} - 16 x + 22$

$8 ((x - 3) (x - 2)) = 0$

$y = 4 x^{2} - 16 x + 22$

Remove unnecessary parentheses.

$8 (x - 3) (x - 2) = 0$

$y = 4 x^{2} - 16 x + 22$

$8 (x - 3) (x - 2) = 0$

$y = 4 x^{2} - 16 x + 22$

$8 (x - 3) (x - 2) = 0$

$y = 4 x^{2} - 16 x + 22$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x - 2 = 0$

$y = 4 x^{2} - 16 x + 22$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$y = 4 x^{2} - 16 x + 22$

Add $3$ to both sides of the equation.

$x = 3$

$y = 4 x^{2} - 16 x + 22$

$x = 3$

$y = 4 x^{2} - 16 x + 22$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = 4 x^{2} - 16 x + 22$

Add $2$ to both sides of the equation.

$x = 2$

$y = 4 x^{2} - 16 x + 22$

$x = 2$

$y = 4 x^{2} - 16 x + 22$

The final solution is all the values that make $8 (x - 3) (x - 2) = 0$ true.

$x = 3, 2$

$y = 4 x^{2} - 16 x + 22$

$x = 3, 2$

$y = 4 x^{2} - 16 x + 22$

Replace all occurrences of $x$ with $3$ in each equation.

Replace all occurrences of $x$ in $y = 4 x^{2} - 16 x + 22$ with $3$ .

$y = 4 {(3)}^{2} - 16 \cdot 3 + 22$

$x = 3$

Simplify the right side.

Simplify $4 {(3)}^{2} - 16 \cdot 3 + 22$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$y = 4 \cdot 9 - 16 \cdot 3 + 22$

$x = 3$

Multiply $4$ by $9$ .

$y = 36 - 16 \cdot 3 + 22$

$x = 3$

Multiply $- 16$ by $3$ .

$y = 36 - 48 + 22$

$x = 3$

$y = 36 - 48 + 22$

$x = 3$

Simplify by adding and subtracting.

Subtract $48$ from $36$ .

$y = - 12 + 22$

$x = 3$

Add $- 12$ and $22$ .

$y = 10$

$x = 3$

$y = 10$

$x = 3$

$y = 10$

$x = 3$

$y = 10$

$x = 3$

$y = 10$

$x = 3$

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = 4 x^{2} - 16 x + 22$ with $2$ .

$y = 4 {(2)}^{2} - 16 \cdot 2 + 22$

$x = 2$

Simplify the right side.

Simplify $4 {(2)}^{2} - 16 \cdot 2 + 22$ .

Simplify each term.

Raise $2$ to the power of $2$ .

$y = 4 \cdot 4 - 16 \cdot 2 + 22$

$x = 2$

Multiply $4$ by $4$ .

$y = 16 - 16 \cdot 2 + 22$

$x = 2$

Multiply $- 16$ by $2$ .

$y = 16 - 32 + 22$

$x = 2$

$y = 16 - 32 + 22$

$x = 2$

Simplify by adding and subtracting.

Subtract $32$ from $16$ .

$y = - 16 + 22$

$x = 2$

Add $- 16$ and $22$ .

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(3, 10)$

$(2, 6)$

The result can be shown in multiple forms.

Point Form:

$(3, 10), (2, 6)$

Equation Form:

$x = 3, y = 10$

$x = 2, y = 6$

Solve by Substitution y+16x-22=4x^2 4x^2-24x+26+y=0

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